Question

A software designer is mapping the streets for a new racing game. All of the streets are depicted as either perpendicular or parallel lines. The equation of the lane passing through A and B is -7x + 3y = -21.5. What is the equation of the central street PQ?

A. -3x + 4y = 3

B. -1.5x − 3.5y = -31.5

C. 2x + y = 20

D. -2.25x + y = -9.75

Answer 1

the given line is

$-7x+3y=-21.5\\ 3y=7x-21.5\\ y=\frac{7}{3}x-\frac{21.5}{3}$

here coefficient of x is 7/3 so the slope of the given line is

$m_1=\frac{7}{3}$

now the line PQ should be either parallel or perpendicular 

let the slope of the line PQ is  m₂

so when m₁=m₂  then both lines are parallel

and when m₁.m₂=-1  then both lines are perpendicular 

now check the first option 

$-3x+4y=3\\ 4y=3x+3\\ y=\frac{3}{4}x+\frac{3}{4}$

here coefficient of x is 3/4 so the slope of the first line is

$m_2= \frac{3}{4} \\ so \space here \space m_1\ne m_2 .$

so both lines are not parallel

now find its product

$m_1 \cdot m_2= \frac{7}{3}\cdot \frac{3}{4}\\ m_1 \cdot m_2= \frac{7}{4}\\ m_1 \cdot m_2 \ne -1\\$

so both lines are not perpendicular 

now check the second option 

$-1.5x-3.5y=-31.5$

$-3.5y=1.5x-31.5$

$3.5y=-1.5x+31.5$

$y=-\frac{1.5}{3.5}x+\frac{31.5}{3.5}$

$y=-\frac{15}{35}x+\frac{315}{35}$

$y=-\frac{3}{7}x+9$

$so\space here \space m_1\ne m_2$

so both lines are not parallel

$now\space find \space its \space product\\ m_1 \cdot m_2= \frac{7}{3}\left(-\frac{3}{7}\right) {\color{Red} m_1 \cdot m_2= -1}$

therefore both lines are perpendicular 

so option B is correct

${\color{Red} -1.5x-3.5y=-31.5} .$